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Full text of "Treatise On Analysis Vol-Ii"

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For if yG A, then
(*,00 I Ug(x)  a) = (U,(x*) ' *,G>) I tf) = W**y) I a) =f(x*y) = (ng(y) \ *,(*)),

and ( follows because A/ng is dense in H^.

We are now in a position to show that g satisfies the condition (N). Let
b e Rg be a vector belonging to the orthogonal supplement of the subspace
generated by the elements ng(xy) for all x, y in A; then we have (ng(xy) \ b) = 0,
that is (Ug(x) - ng(y) \ b) = 0, or again (ng(y) \ Ug(x*) - b) = 0 for all x and y in
A. Since the elements ng(y) form a dense subspace of Hg, it follows that
(a| Ug(x*)  6) = 0, hence (Ug(x) -a\b) = Q9 hence finally (ng(x) \b) = 0 and
therefore 6 = 0, because A/ng is dense in Hff.

Now consider the positive linear form/"(F) = (V-a\a) on the algebra
j*'g. We have


by virtue of the Gelfand-Neumark theorem (15.4.14), we may write/"(F) =
h(<SV) where h is a linear form on ^C(X(^)), and since ||^F|| = \\V\\, it
follows that h is an (a priori complex) measure on the compact space X(^).
Since any continuous function G g; 0 on X(^) is of the form F  F, hence
of the form 9V* &V* =^(KF*), we have h(G) = || F- cr||2 ^ 0, so that A is a
positive measure. Taking into account (, ( and the canonical
homeomorphism co of X(^) onto S^, we see therefore that there exists a
positive measure v on $g (induced by the measure co(h) on S^, and therefore
bounded) such that


for all x e A.

Thus it remains to show that v = ma. Since


it is sufficient, by virtue of (15.9.2(11)) to show that (1) the functions x (where
x e A) belong to J^c(Sff, v); (2) the support of v is Sff. The first assertion is
trivial, because the functions x are bounded and continuous and the measure
v is bounded. To prove (2), suppose that there exists a function F ^ 0 belong-
ing to jf c(Sff) such that jF(x) dv(x) = 0. Then JF(x)x(x)J^) dv(X) = 0 for all
x, ye A. Since F o co is a continuous function on X(sfg), it is of the form
^F where Ve$0rg, and the relation above now takes the formhat the set of functions x is dense in JSfc(S, m), and and therefore definesions xnxn therefore converge uniformly on S^ to G, and hence,